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Shiroha白羽的博客

Golang 踩坑 —— interface 为参数的时候传 nil 指针 Codeforces Round 925 (Div. 3) Codeforces Round 924 (Div. 2) Codeforces Round 923 (Div. 3) Codeforces Round 922 (Div. 2) Codeforces Round 921 (Div. 2) Educational Codeforces Round 161 (Rated for Div. 2) Codeforces Round 920 (Div. 3) Codeforces Round 919 (Div. 2) Hello 2024 Good Bye 2023 Codeforces Round 918 (Div. 4) 个人备份的常用 macOS 清理命令 Codeforces Round 917 (Div. 2) Pinely Round 3 (Div. 1 + Div. 2) Educational Codeforces Round 160 (Rated for Div. 2) Codeforces Round 915 (Div. 2) Codeforces Round 914 (Div. 2) Codeforces Round 913 (Div. 3) Educational Codeforces Round 159 (Rated for Div. 2) Codeforces Round 912 (Div. 2) Codeforces Round 911 (Div. 2) CodeTON Round 7 (Div. 1 + Div. 2, Rated, Prizes!) Educational Codeforces Round 158 (Rated for Div. 2) Codeforces Round 910 (Div. 2) Codeforces Round 909 (Div. 3) Codeforces Round 908 (Div. 2) Educational Codeforces Round 157 (Rated for Div. 2) C++自定义的字面量 Codeforces Round 907 (Div. 2)
2020牛客暑期多校训练营(第一场)H-Minimum-cost Flow——网络流
Shiroha · 2020-07-14 · via Shiroha白羽的博客

给出一个费用流图,每条边的流量上限相同且不固定。有$q$个询问,每个询问中给出每条边的流量上限(分数,且保证$\leq 1$)。当图中的流量为 $1$ 个单位的时候,求出此时的费用。

首先由于边的流量均为分数($\frac{u}{v}$),而总流量为 $1$ 个单位。我们先扩大$\frac{v}{u}$倍,将每条边的流量固定为 $1$ 个单位,此时流量为 $\frac{v}{u}$ 个单位

考虑在最大流使用 SPFA 查找路径时,当查找到一条路径时,此路径的流量一定为 $1$ (根据上述的设定)。由于采用的本身便是最短路查找路径,得到的路径的费用为当前网络图下的最低费用。

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#define ll long long

const int maxn = 100; //点数
const int INF = 0x3f3f3f3f;

struct Edge {
int from, to, cap, flow, cost;

Edge(int u, int v, int c, int f, int cc)
: from(u), to(v), cap(c), flow(f), cost(cc) {}
};

vector<ll> res;

struct MCMF {
int n, m;
vector<Edge> edges;
vector<int> G[maxn];
int inq[maxn]; //是否在队列中
int d[maxn]; //bellmanford
int p[maxn]; //上一条弧
int a[maxn]; //可改进量
void init(int n) {
this->n = n;
for (int i = 0; i <= n; ++i) G[i].clear();
edges.clear();
}

void addEdge(int from, int to, int cap, int cost) {
edges.emplace_back(Edge(from, to, cap, 0, cost));
edges.emplace_back(Edge(to, from, 0, 0, -cost));
m = int(edges.size());
G[from].emplace_back(m - 2);
G[to].emplace_back(m - 1);
}

bool spfa(int s, int t, int &flow, ll &cost) {
for (int i = 1; i <= n; ++i) d[i] = INF;
memset(inq, 0, sizeof(inq));
d[s] = 0;
inq[s] = 1;
p[s] = 0;
queue<int> q;
a[s] = INF;
q.push(s);
while (!q.empty()) {
int u = q.front();
q.pop();
inq[u] = 0;
for (int i = 0; i < int(G[u].size()); ++i) {
Edge &e = edges[G[u][i]];
if (e.cap > e.flow && d[e.to] > d[u] + e.cost) {
d[e.to] = d[u] + e.cost;
p[e.to] = G[u][i];
a[e.to] = min(a[u], e.cap - e.flow);
if (!inq[e.to]) {
q.push(e.to);
inq[e.to] = 1;
}
}
}
}
if (d[t] == INF) return false;
flow += a[t];
cost += (ll) d[t] * (ll) a[t];
res.push_back(d[t]);
for (int u = t; u != s; u = edges[p[u]].from) {
edges[p[u]].flow += a[t];
edges[p[u] ^ 1].flow -= a[t];
}
return true;
}

int MincostMaxflow(int s, int t, ll &cost) {
int flow = 0;
cost = 0;
while (spfa(s, t, flow, cost));
return flow;
}
} mcmf;

虽然 $1 \leq u, v \leq 1e9$,可能需要循环遍历 $1e9$ 次才能出结果,但是由于图中每条边的容量都是 $1$,所以对于每一条路径,它只有被完全占用和完全空闲两种状态,而边数只有 $100$ 条,即整个网络的最大流量只能是 $100$ 个单位,即最多遍历 $100$ 次,则复杂度不超过 $100 1e5 = 1e7$,当 $MAX_FLOW u < v$ 时,则无解,输出 $-1$

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#include <bits/stdc++.h>

using namespace std;

#define ll long long

const int maxn = 100; //点数
const int INF = 0x3f3f3f3f;

struct Edge {
int from, to, cap, flow, cost;

Edge(int u, int v, int c, int f, int cc)
: from(u), to(v), cap(c), flow(f), cost(cc) {}
};

vector<long long> res;

struct MCMF {
int n, m;
vector<Edge> edges;
vector<int> G[maxn];
int inq[maxn]; //是否在队列中
int d[maxn]; //bellmanford
int p[maxn]; //上一条弧
int a[maxn]; //可改进量
void init(int n) {
this->n = n;
for (int i = 0; i <= n; ++i) G[i].clear();
edges.clear();
}

void addEdge(int from, int to, int cap, int cost) {
edges.emplace_back(Edge(from, to, cap, 0, cost));
edges.emplace_back(Edge(to, from, 0, 0, -cost));
m = int(edges.size());
G[from].emplace_back(m - 2);
G[to].emplace_back(m - 1);
}

bool spfa(int s, int t, int &flow, ll &cost) {
for (int i = 1; i <= n; ++i) d[i] = INF;
memset(inq, 0, sizeof(inq));
d[s] = 0;
inq[s] = 1;
p[s] = 0;
queue<int> q;
a[s] = INF;
q.push(s);
while (!q.empty()) {
int u = q.front();
q.pop();
inq[u] = 0;
for (int i = 0; i < int(G[u].size()); ++i) {
Edge &e = edges[G[u][i]];
if (e.cap > e.flow && d[e.to] > d[u] + e.cost) {
d[e.to] = d[u] + e.cost;
p[e.to] = G[u][i];
a[e.to] = min(a[u], e.cap - e.flow);
if (!inq[e.to]) {
q.push(e.to);
inq[e.to] = 1;
}
}
}
}
if (d[t] == INF) return false;
flow += a[t];
cost += (ll) d[t] * (ll) a[t];
res.push_back(d[t]);
for (int u = t; u != s; u = edges[p[u]].from) {
edges[p[u]].flow += a[t];
edges[p[u] ^ 1].flow -= a[t];
}
return true;
}

int MincostMaxflow(int s, int t, ll &cost) {
int flow = 0;
cost = 0;
while (spfa(s, t, flow, cost));
return flow;
}
} mcmf;

int n, m;

void solve() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.precision(10);
cout << fixed;
while (cin >> n >> m) {
mcmf.init(n + 10);
res.clear();
for (int i = 0; i < m; ++i) {
int u, v, f;
cin >> u >> v >> f;
mcmf.addEdge(u, v, 1, f);
}
ll cost = 0;
ll mf = mcmf.MincostMaxflow(1, n, cost);
int q;
cin >> q;
while (q--) {
ll u, v;
cin >> u >> v;
if (mf * u < v) {
cout << "NaN" << '\n';
continue;
}
ll sum = 0;
ll ans = 0;
for (auto item : res) {
if (sum + u <= v) {
sum += u;
ans += item * u;
} else {
ans += (v - sum) * item;
break;
}
}
ll g = __gcd(ans, v);
cout << ans / g << '/' << v / g << '\n';
}
}
}

signed main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
#ifdef ACM_LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
int test_index_for_debug = 1;
char acm_local_for_debug;
while (cin >> acm_local_for_debug) {
if (acm_local_for_debug == '$') exit(0);
cin.putback(acm_local_for_debug);
if (test_index_for_debug > 20) {
throw runtime_error("Check the stdin!!!");
}
auto start_clock_for_debug = clock();
solve();
auto end_clock_for_debug = clock();
cout << "Test " << test_index_for_debug << " successful" << endl;
cerr << "Test " << test_index_for_debug++ << " Run Time: "
<< double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
cout << "--------------------------------------------------" << endl;
}
#else
solve();
#endif
return 0;
}